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I am struggling to factor a larger expression. I seem to have forgotten how to do it, hence I am coming here for help, so that I can refresh the concept.

The equation is $$x^3+2x+3=0$$

How do solve it by splitting the middle term? I tried the total and error method, and subsequent long division to obtain $$(x+1)(x^2-x+3)$$

So I have indeed found the factors, but how to do it without lengthy calculations? The middle term can be split, I have just forgotten how to.

Edit: I see that splitting the middle term is the wrong term. Forgive me.

What I meant was using formulas like $a^3+b^3=(a+b)(a^2-ab+b^2)$

I am fairly confident that’s how it’s done, because I have used this before.

It’s extremely embarrassing to ask this, but it had to be done.

Aditya
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    "The middle term can be split" Uhm... are you sure you aren't misremembering techniques learned for the purposes of factoring quadratics which don't apply to factoring cubics? There does exist a general approach for cubics, but people very infrequently learn it beyond having learned about its existence, and fewer still use it in practice. – JMoravitz Feb 13 '20 at 15:24
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    The most generalized approach to factoring a cubic that is commonly used is called Cardano's Formula. This Mathologer video is well worth watching from the beginning for anyone even mildly interested in this, but if you just want a quick answer you might just care about watching from where I timestamped. Reading about it instead at proof wiki can also help. – JMoravitz Feb 13 '20 at 15:28
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    Most commonly, when encountering a problem like this, Cardano's Formula is not what is intended to be used (though works wonders if you can master it). What is instead commonly intended is checking to see if $0,1,-1,2,-2$ or other quickly identifiable numbers are a root, and then perform long division as you had. – JMoravitz Feb 13 '20 at 15:30
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    @JMoravitz Better yet, knowing the rational root theorem, only checking the divisors of the constant term $3$. – Servaes Feb 13 '20 at 15:30
  • @JMoravitz please check the edit – Aditya Feb 13 '20 at 15:54

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When I was at university, we did this kind of factorisation thru the rational root theorem.

If $\frac pq$ is a rational root of a polynomial (with integer coefficients), then $p$ divide the constant term and $q$ divide the main coefficient.

In your example $$x^3+2x+3$$ we must have $p|3$ and $q|1$. We only have to check for $\pm1$ and $\pm3$.

You asked if there was a way to split the middle terms (there are two middle terms). We could develop something similar to quadratic factorisation. Let's work backward. $$(x+1)(x^2-x+3)$$ We distribute the first factor thru the second $$x^2(x+1)-x(x+1)+3(x+1)\tag{1}$$ distribute again $$x^3+x^2-x^2-x+3x+3\tag{2}$$ If we want to split the middle terms, we need to find equation $(2)$.

Starting with your polynomial $$x^3+2x+3$$ We are looking for integers $A$, $B$, $C$ and $D$ to split the middle term like this $$x^3+Ax^2+Bx^2+Cx+Dx+3$$ with conditions $$A+B=0\tag{Cond1}$$ $$C+D=2\tag{Cond2}$$ Futhermore, we want to be able to factorise it to look like equation $(1)$. With the first two terms, there is a $x^2$ in common $$x^2(x+A)+Bx^2+Cx+Dx+3$$ We want to have the exact same parenthesis with the other four terms, so we need $$x^2(x+A)+Bx\left(x+\tfrac CB\right)+D\left(x+\tfrac3D\right)$$ For the parenthesis to be the same, we need $$A=\frac CB\tag{Cond3}$$ $$A=\frac3D\tag{Cond4}$$ From $(\text{Cond}4)$, $D=\pm1$ or $D=\pm3$. From $(\text{Cond}1)$, $A=-B$, combined with $(\text{Cond}3)$, $C=-A^2$. We now try a value of $D$, which force the other value, finally check if it fits in $(\text{Cond}2)$

The one that work is $D=3$, then $A=1$, $C=-1$ and $B=-1$. It is easy to see $C+D=2$. $$x^3+Ax^2+Bx^2+Cx+Dx+3$$ $$x^3+x^2-x^2-x+3x+3$$ $$x^2(x+1)-x(x+1)+3(x+1)$$ $$(x+1)(x^2-x+3)$$

It work realatively well since the polynomial is simple. It could be a lot more complicated if the leading coefficient was different of $1$.

Alain Remillard
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  • How did you establish the conditions? A+B=0 known only because we knew what the final equation was going to be – Aditya Feb 13 '20 at 16:48
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    We start with your polynomial $x^3+2x+3$. Since it doesn't have a term in $x^2$, then $Ax^2+Bx^2$ must be $0$. Thus $A+B=0$. – Alain Remillard Feb 13 '20 at 16:55